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EE70 Review
Electrical Current
dq (t )
i (t ) 
dt
t
q (t )   i (t )dt  q (t0 )
t0
Circuit Elements
An electrical circuit consists of circuit elements such as voltage sources,
resistances, inductances and capacitances that are connected in closed
paths by conductors
Reference Directions
The voltage vab has a reference polarity that is positive
at point a and negative at point b.
Reference Directions
Reference Directions
“uphill: battery”
“downhill: resistor”
Energy is transferred when charge flows through an
element having a voltage across it.
Power and Energy
p(t )  v(t )i (t )
Watts
t2
w   p(t )dt
t1
Joules
Reference Direction
Current is flowing in
the passive
configuration
• If current flows in the passive configuration the
power is given by p = vi
• If the current flows opposite to the passive
configuration, the power is given by p = -vi
Dependent Sources
Resistors and Ohm’s Law
a
b
v  iR
vab  iab R
The units of resistance are Volts/Amp which are called
“ohms”. The symbol for ohms is omega: 
Resistance Related to Physical
Parameters
R
L
A
 is the resistivity of the material used to fabricate
the resistor. The units of resitivity are ohm-meters
(-m)
Power dissipation in a resistor
2
v
p  vi  Ri 
R
2
Kircohoff’s Current Law (KCL)
• The net current entering a node is zero
• Alternatively, the sum of the currents
entering a node equals the sum of the
currents leaving a node
Kircohoff’s Current Law (KCL)
Series Connection
ia  ib i c
Kircohoff’s Voltage Law (KVL)
The algebraic sum of the voltages
equals zero for any closed path (loop) in
an electrical circuit.
Kircohoff’s Voltage Law (KVL)
Parallel Connection
KVL through A and B: -va+vb = 0  va = vb
KVL through A and C: -va - vc = 0  va = -vc
Equivalent Series Resistance
v  v1  v2  v3  iR1  iR2  iR3  i( R1  R2  R3 )  iReq
Equivalent Parallel Resistance
 1
v
v
v
1
1 
v


i  i1  i2  i3 


 v




R1 R2 R3
 R1 R2 R3  Req
Circuit Analysis using
Series/Parallel Equivalents
1. Begin by locating a combination of
resistances that are in series or
parallel. Often the place to start is
farthest from the source.
2. Redraw the circuit with the equivalent
resistance for the combination found in
step 1.
Voltage Division
R1
v1  R1i 
v total
R1  R2  R3
R2
v2  R2 i 
v total
R1  R2  R3
Of the total voltage, the fraction that appears across a
given resistance in a series circuit is the ratio of the given
resistance to the total series resistance.
Current Division
R2
v
i1 

itotal
R1 R1  R2
R1
v
i2 

itotal
R2 R1  R2
For two resistances in parallel, the fraction of the total
current flowing in a resistance is the ratio of the other
resistance to the sum of the two resistances.
Node Voltage Analysis
Node Voltage Analysis
v1  v s
v2  v1 v2 v2  v3


0
R2
R4
R3
v3  v1 v3 v3  v2


0
R1
R5
R3
Mesh Current Analysis
Mesh Current Analysis
Thévenin Equivalent Circuits
Thévenin Equivalent Circuits
Vt  voc
Thévenin Equivalent Circuits
Vt
isc 
Rt
Thévenin Equivalent Circuits
voc
Rt 
isc
Thévenin Equivalent Circuits
Norton Equivalent Circuits
Norton Equivalent Circuits
I n  isc
Source Transformations
Maximum Power Transfer
Vt
IL 
Rt  RL
PL  iL2 RL
 Vt
 
 Rt  RL
2

 RL


dPL
 0  RL  Rt
dRL
Superposition Principle
The superposition principle states
that the total response is the sum of
the responses to each of the
independent sources acting
individually. In equation form, this is
rT  r1  r2    rn
Superposition Principle
Superposition Principle
Current source
open circuit
R2
5
v1 
vs 
15V  5V
R1  R2
5  10
Superposition Principle
Voltage source
short circuit
v2  is Req
Req
R1 R2
(10 )(5)
 is
 (2 A)
 (2 A)(3.33)  6.66V
R1  R2
10  5
vT  v1  v2  5V  6.66V  11 .66V
Voltage-Amplifier Model
The input resistance Ri is the equivalent resistance see when
looking into the input terminals of the amplifier. Ro is the output
resistance. Avoc is the open circuit voltage gain.
Voltage Gain
Ideally, an amplifier produces an output
signal with identical waveshape as the
input signal, but with a larger amplitude.
vo t   Av vi t 
Current Gain
io
Ai 
ii
io vo RL
Ri
Ai  
 Av
ii
vi Ri
RL
Power Gain
Po
G
Pi
Po Vo I o
2 Ri
G

 Av Ai   Av 
Pi Vi I i
RL
Operational Amplifier
Summing Point Constraint
Operational amplifiers are almost always
used with negative feedback, in which part of
the output signal is returned to the input in
opposition to the source signal.
Summing Point Constraint
In a negative feedback system, the ideal opamp output voltage attains the value needed
to force the differential input voltage and input
current to zero. We call this fact the
summing-point constraint.
Summing Point Constraint
1. Verify that negative feedback is present.
2. Assume that the differential input voltage
and the input current of the op amp are
forced to zero. (This is the summing-point
constraint.)
3. Apply standard circuit-analysis principles,
such as Kirchhoff’s laws and Ohm’s law, to
solve for the quantities of interest.
The Basic Inverter
Applying the Summing Point Constraint
Inverting Amplifier
vo
R2
Av

vin
R1
Summing Amplifier
Non-inverting Amplifiers
vo
R2
Av
1
vin
R1
Voltage Follower
vo
R2
0
A v
 1
 1  1
vin
R1

Capacitance
q
C
v
qt  
q
v
C
t

it dt  qt 0 
t0
t
1
v t    i t dt  v t0 
C t0
Capacitances in Parallel
Capacitances in Series
1
1
1
1



Ceq C1 C 2 C3
Inductance
Inductance
The polarity of the voltage is such as to oppose the change in
current (Lenz’s law).
Series Inductances
Leq  L1  L2  L3
Parallel Inductances
1
1

Leq L1  L2  L3
Mutual Inductance
Fields are aiding
Fields are opposing
Magnetic flux produced by one coil links the other coil
Discharge of a Capacitance
through a Resistance
iC
iR
KCL at the top node of the
circuit:
q
dq
dv
C   q  Cv  iC 
C
v
dt
dt
vC (t )
iR 
R
dvC t  vC t 
C

0
dt
R
dvC t 
RC
 vC t   0
dt
Discharge of a Capacitance through a
Resistance
dvC t 
RC
 vC t   0
dt
dvC t 
1

vC t 
dt
RC
We need a function vC(t) that has the same form as it’s
derivative.
vC t   Ke
st
Substituting this in for vc(t)
RCKse  Ke  0
st
st
Discharge of a Capacitance through a
Resistance
1
Solving for s: s 
RC
Substituting into vc(t):
Initial Condition:
vC 0    Vi
vC t   Ke
t RC
Full Solution:
vC t   Vi e
t RC
Discharge of a Capacitance through a
Resistance
vC t   Ke
t RC
To find the unknown constant K, we need to use the
boundary conditions at t=0. At t=0 the capacitor is initially
charged to a voltage Vi and then discharges through the
resistor.
vC 0    Vi
vC t   Vi e
t RC
Discharge of a Capacitance through a
Resistance
Charging a Capacitance from a DC
Source through a Resistance
Charging a Capacitance from a DC
Source through a Resistance
KCL at the node that joins
the resistor and the capacitor
dvC (t ) vC (t )  VS
C

0
dt
R
Current into the
dvc
capacitor:
C
dt
Current through the
resistor: vC (t )  VS
R
Charging a Capacitance from a DC
Source through a Resistance
dvC (t ) vC (t )  VS
C

0
dt
R
Rearranging:
dvC (t )
RC
 vC (t )  VS
dt
This is a linear first-order differential equation with
contant coefficients.
Charging a Capacitance from a DC
Source through a Resistance
The boundary conditions are given by the fact that the
voltage across the capacitance cannot change
instantaneously:
vC (0)  vC (0)  0
Charging a Capacitance from a DC
Source through a Resistance
Try the solution:
vC (t )  K1  K 2e st
Substituting into the differential equation:
Gives:
dvC (t )
RC
 vC (t )  VS
dt
(1  RCs) K 2e st  K1  VS
Charging a Capacitance from a DC
Source through a Resistance
(1  RCs) K 2e  K1  VS
st
For equality, the coefficient of est must be zero:
1
1  RCs  0  s 
RC
Which gives K1=VS
Charging a Capacitance from a DC
Source through a Resistance
Substituting in for K1 and s:
vC (t )  K1  K 2e st  VS  K 2e t / RC
Evaluating at t=0 and remembering that vC(0+)=0
vC (0)  VS  K 2e0  VS  K 2  0  K 2  Vs
Substituting in for K2 gives:
vC (t )  K1  K 2e st  VS  VS e t / RC
Charging a Capacitance from a DC
Source through a Resistance
vC t   Vs  Vs e
t 
DC Steady State
dvC (t )
iC (t )  C
dt
In steady state, the voltage is constant, so the
current through the capacitor is zero, so it behaves
as an open circuit.
DC Steady State
diL (t )
v L (t )  L
dt
In steady state, the current is constant, so the
voltage across and inductor is zero, so it behaves
as a short circuit.
DC Steady State
The steps in determining the forced response for
RLC circuits with dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.
RL Transient Analysis
 VS  i(t ) R  v(t )  0
di (t )
i(t ) R  L
 VS
dt
RL Transient Analysis
i (t ) R  L
di (t )
 VS
dt
di(t )
i (t ) R  L
 VS
dt
Try i(t )  K1  K 2e st
Try i(t )  K1  K 2e st
RK 1  ( RK 2  sLK 2 )e  VS
st
VS 100V
RK 1  VS  K1 

 2A
R 50
L
RK 2  sLK 2  0  s 
R
RC and RL Circuits with General
Sources
First order differential
equation with constant
coefficients
di (t )
L
 Ri (t )  vt (t )
dt
vt (t )
L di(t )
 i (t ) 
R dt
R
dx(t )
Forcing

 x(t )  f (t ) function
dt
RC and RL Circuits with General
Sources
The general solution consists
of two parts.
The particular solution (also called the
forced response) is any expression that
satisfies the equation.
dx(t )

 x(t )  f (t )
dt
In order to have a solution that satisfies the
initial conditions, we must add the
complementary solution to the particular
solution.
The homogeneous equation is obtained by
setting the forcing function to zero.
dx(t )

 x(t )  0
dt
The complementary solution (also called
the natural response) is obtained by
solving the homogeneous equation.
Integrators and Differentiators
Integrators produce output voltages that are
proportional to the running time integral of
the input voltages. In a running time integral,
the upper limit of integration is t .
1
vo t   
RC
t

0
vin t dt
Differentiator Circuit
dvin
vo t    RC
dt
Second–Order Circuits
di t 
1
L
 Ri t    i t dt  vC 0  vs t 
dt
C0
t
Differentiating with respect to time:
2
L
d i(t )
dt
2
dvs (t )
di(t ) 1
R
 i(t ) 
dt
C
dt
Second–Order Circuits
d 2i(t )
dt
Define:
2
R di(t )
1
1 dvs (t )


i(t ) 
L dt
LC
L dt
R

2L
0 
1
LC
Dampening
coefficient
Undamped resonant
frequency
1 dvs (t )
f (t ) 
Forcing function
L dt
2
d i(t )
dt
2
di(t )
2
 2
 0 i(t )  f (t )
dt
Solution of the Second-Order Equation
Particular solution
d it 
dit 
2





2



i
t

f
t
0
2
dt
dt
Complementary solution
2
d it 
2
dt
2
dit 
2
 2
 0 it   0
dt
Solution of the Complementary
Equation
d it 
2
dt 2
dit 
 2
 02it   0
dt
Try iC (t )  Ke st :
s 2 Ke st  2sKe st  02 Ke st  0
Factoring :
( s 2  2s  02 ) Ke st  0
Characteristic equation:
s 2  2s  02  0
Solution of the Complementary
Equation
Roots of the characteristic equation:
s1      
2
s2      
2

 
0
2
0
2
0
Dampening ratio
1. Overdamped case (ζ > 1). If ζ > 1 (or
equivalently, if α > ω0), the roots of the
characteristic equation are real and distinct.
Then the complementary solution is:
ic t   K1e
s1t
 K 2e
s 2t
In this case, we say that the circuit is
overdamped.
2. Critically damped case (ζ = 1). If ζ = 1 (or
equivalently, if α = ω0 ), the roots are real
and equal. Then the complementary solution
is
ic t   K1e
s1t
 K 2te
s1t
In this case, we say that the circuit is
critically damped.
3. Underdamped case (ζ < 1). Finally, if ζ < 1
(or equivalently, if α < ω0), the roots are
complex. (By the term complex, we mean that
the roots involve the square root of –1.) In other
words, the roots are of the form:
s1    jn and s2    jn
in which j is the square root of -1 and the
natural frequency is given by:
n    
2
0
2
For complex roots, the complementary solution
is of the form:
ic t   K1e
t
cosnt   K 2e
t
sinnt 
In this case, we say that the circuit is
underdamped.
Circuits with Parallel L and C
V
iR 
R
t

1
dq
dv
iL 
vdt  i L (0) ic 
C
L
dt
dt
0
We can replace the circuit with it’s Norton equivalent
and then analyze the circuit by writing KCL at the top
node:
t
C

dv(t ) 1
1
 v(t ) 
v(t )dt  i L (0)  in (t )
dt
R
L
0
Circuits with Parallel L and C
t

dv(t ) 1
1
C
 v(t ) 
v(t )dt  iL (0)  in (t )
dt
R
L
0
differentiating :
din (t )
d 2 v(t ) 1 dv(t ) 1
C

 v(t ) 
dt
R dt
L
dt
d 2 v(t )
1 dv(t )
1
1 din (t )


v(t ) 
dt
RC dt
LC
C dt
Circuits with Parallel L and C
d 2 v(t )
1 dv(t )
1
1 din (t )


v(t ) 
dt
RC dt
LC
C dt
1
Dampening coefficient  
2 RC
1
Undamped resonant frequency  0 
LC
1 din (t )
Forcing function f (t ) 
C dt
d 2 v(t )
dv(t )
2
 2
  0 v(t )  f (t )
dt
dt
Circuits with Parallel L and C
d 2 v(t )
dv(t )
 2
 02 v(t )  f (t )
dt
dt
This is the same equation as we found for the series LC
circuit with the following changes for :
1
Parallel circuit  
2 RC
R
Series circuit  
2L
Complex Impedances-Inductor
VL  jL  I L
Z L  jL  L90
VL  Z L I L

Complex Impedances-Inductor
Complex Impedances-Capacitor
VC  Z C I C
1
1
1

ZC   j


  90
C jC C
VR  RI R
Complex Impedances-Capacitor
Impedances-Resistor
VR  RI R
Impedances-Resistor
Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors.
The sum of the phasor voltages equals
zero for any closed path.
V1  V2  V3  0
The sum of the phasor currents entering a
node must equal the sum of the phasor
currents leaving.
I in  I out
Power in AC Circuits
V Vm 0
I 
 I m  
Z
Z 

Vm
where I m 
Z
For >0 the load is called “inductive” since Z=jL for an inductor
For <0 the load is “capacitive” since Z=-j/C for a capacitor
Load Impedance in the Complex Plane
Z  Z   R  jX
R
cos( ) 
Z
X
sin(  ) 
Z
Power for a General Load
If the phase angle for the voltage is not zero, we
define the power angle :
Power angle:
   voltage   current
P  Vrms I rms cos( )
PF  cos( )
AC Power Calculations
Average Power: P  Vrms I rms cos  W 
Reactive Power: Q  Vrms I rms sin   VAR
Apparent Power: VRMS I RMS
VA
Power Triangles
Average power
P  Q  Vrms I rms 
2
Average
power
2
2
Reactive
power
Apparent
power
Thevenin Equivalent Circuits
The Thevenin equivalent for an ac circuit consists of a
phasor voltage source Vt in series with a complex
impedance Zt
Thevenin Equivalent Circuits
The Thévenin voltage is equal to the open-circuit
phasor voltage of the original circuit.
Vt  Voc
We can find the Thévenin impedance by zeroing the
independent sources and determining the impedance
looking into the circuit terminals.
Thevenin Equivalent Circuits
The Thévenin impedance equals the open-circuit
voltage divided by the short-circuit current.
Voc Vt
Z t

I sc I sc
Norton Equivalent Circuit
The Norton equivalent for an ac circuit consists of a
phasor current source In in parallel with a complex
impedance Zt
I n  I sc
Maximum Average Power Transfer
The Thevenin equivalent of a two-terminal circuit
delivering power to a load impedance.
Maximum Average Power Transfer
If the load can take on any complex value,
maximum power transfer is attained for a load
impedance equal to the complex conjugate of the
Thévenin impedance.
If the load is required to be a pure resistance,
maximum power transfer is attained for a load
resistance equal to the magnitude of the
Thévenin impedance.
Transfer Functions
The transfer function H(f ) of the two-port filter
is defined to be the ratio of the phasor output
voltage to the phasor input voltage as a function
of frequency:
Vout
Hf  
Vin
First-Order Low Pass Filter
Vin
I
1
R
j 2fC




 1 
Vin
1


Vout 
I  
j 2fC
 j 2fC  R  1 


j
2

fC


V
1
1
H ( f )  out 

Vin 1  j 2fRC 1  j ( f / f B )
fB 
1
2RC
Half power
frequency
First-Order Low Pass Filter
1
H f 

1  j f f B 
H f  
1
1   f fB 
2
10
1  f fB 
2
 f 

 arctan
 fB 
 f 
H  f    arctan  
 fB 
First-Order Low Pass Filter
For low frequency signals the magnitude of the transfer
function is unity and the phase is 0. Low frequency signals
are passed while high frequency signals are attenuated and
phase shifted.
Magnitude Bode Plot for First-Order
Low Pass Filter
1. A horizontal line at zero for f < fB /10.
2. A sloping line from zero phase at fB /10 to –90° at 10fB.
3. A horizontal line at –90° for f > 10fB.
First-Order High-Pass Filter
Vout 
Vout
Vin
R
Vin 
1
R j
2fC
j( f / f B )

1  j( f / f B )
1
1
1 j
2fRC
j 2fRC
Vin 
Vin
j 2fRC  1
1
where f B 
2fRC
First-Order High-Pass Filter
Vout

j f f B 
f f B 90 
Hf 


Vin 1  j  f f B 
1   f f B 2  arctan f f B 
Hf  
 f fB 
1   f f B 2
90 
H  f  
 90   arctan f f B 
 arctan f f B 
First-Order High-Pass Filter
H f  
 f fB 
1   f f B 2
H  f   90  arctan f f B 

Bode Plots for the First-Order HighPass Filter

H  f   20 log( f f B )  10 log 1   f f B 2
For f  f B H  f   20 log( f f B )
For f  f B H  f   20 log( f f B )
For f  f B H  f   20 log( f f B )  10 log  f f B 2  0

For f  f B H  f   20 log( f f B )  10 log  f f B 2  0
 f 

H  f   90   arctan
 fB 
For f  f B H  f   90 
For f  f B H  f   0 
Series Resonance
Z s ( f )  j 2fL  R  j
1
2fC
resonanceof
: the inductor and the capacitor cancel:
For resonance For
the reactance
1
1
2
2f 0 L 
 f0 
2f 0 C
(2 ) 2 LC
1
f0 
2 LC
Series Resonance
Quality factor QS
Qs 

Substitute L 
Qs 
Reactance of inductanceat resonance
Resistance
2f 0 L
R
1
1
from f 0 
2
2
(2 ) ( f 0 ) C
2 LC
1
2f 0 CR
Series Resonant Band-Pass Filter
Vs
I

Zs ( f )
Vs / R
 f
f0 
1  jQ s   
f 
 f0
Vs
V
1
VR  RI 
 R 
Vs
 f
 f
f0 
f0 
1  jQ s   
1  jQ s   
f 
f 
 f0
 f0
Series Resonant Band-Pass Filter
VR
1

Vs 1  jQs  f f 0  f 0 f 
Parallel Resonance
1
Zp 
1 R   j 2fC  j 1 2fL
At resonance ZP is purely resistive:
j 2f 0C  j 1 2f 0 L   f 0 
1
2 LC
Parallel Resonance
Quality factor QP


Resistance

QP  
 Reactance of inductanceat resonance 
R

2f 0 L
Substitute L 
1
(2 ) ( f 0 ) C
2
QP  2f 0CR
2
from f 0 
1
2 LC
Parallel Resonance
Vout
IR
 IZ P 
 f
f0 
1  jQ P   
f 
 f0
Vout for constant current,
varying the frequency
Second Order Low-Pass Filter
j
2fC
Vout
j
ZC
2fRC

Vin 
Vin 
Vin
j
Z R  Z L  ZC

2f 0 L  f
1
R  j 2fL 
 

1 j
2fC
R  f 0 2ff 0 LC 
Vout
Vin
j
 jQ S ( f / f 0 )
2fRC
 H( f ) 


 f
2f 0 L  f
f 
1
 
 1  jQ S   0 
1 j
R  f 0 2ff 0 LC 
f 
 f0
Second Order Low-Pass Filter
Vout
 jQ s  f 0 f 
Hf 

Vin
1  jQ s  f f 0  f 0 f 

Qs  f 0 f   90 
2
1  QS
Hf  
f
f 0  f 0 f  Tan Qs  f f 0  f 0 f 
1
2
Qs  f 0 f 
2
1  QS
f
f0  f0 f 
2
Second Order Low-Pass Filter
Second Order High-Pass Filter
At low frequency the capacitor
is an open circuit
At high frequency the
capacitor is a short and the
inductor is open
Second Order Band-Pass Filter
At low frequency the capacitor
is an open circuit
At high frequency the inductor
is an open circuit
Second Order Band-Reject Filter
At low frequency the
capacitor is an open
circuit
At high frequency the
inductor is an open
circuit
First-Order Low-Pass Filter
Zf
Vo
H( f ) 

Vi
Zi
1
1


Zf
Rf
Zf 
j 2fR f C f
1
1


1
Rf
Rf
j 2fC f
Rf
1  j 2fR f C f
 Rf 
1

H( f )  
 

Zi
 Ri  1  j 2fR f C f
 R f 

1

 
 1  j( f / f ) 
R
B 
 i 
1
fB 
2R f C f
Zf
A low-pass filter with a dc gain of -Rf/Ri
First-Order High-Pass Filter
Zf
vo
H( f ) 

vi
Zi
Z i  Ri 
H( f )  
Zf
Zi
1
j 2 f Ci
Z f  Rf
Rf

Ri 
j 2 f R f Ci
1
j 2 f Ci
R
  i
1  j 2 f Ri Ci
 Ri
 R f  j ( f / f B ) 

 
 1  j( f / f ) 
R
i
B 


1
fB 
2 Ri Ci

 j 2 f R f Ci

 1  j 2 f R C
i i

A high-pass filter with a high frequency gain of -Rf/Ri
Flux Linkages and Faraday’s Law
Magnetic flux passing through a surface area A:


B  dA
A
For a constant magnetic flux density perpendicular
to the surface:
 BA
The flux linking a coil with N turns:
  N
Faraday’s Law
Faraday’s law of magnetic induction:
d
e
dt
The voltage induced in a coil whenever its
flux linkages are changing. Changes occur
from:
• Magnetic field changing in time
• Coil moving relative to magnetic field
Lenz’s Law
Lenz’s law states that the polarity of the
induced voltage is such that the voltage
would produce a current (through an
external resistance) that opposes the
original change in flux linkages.
Lenz’s Law
Magnetic Field Intensity and Ampère’s
Law
B  H H  Magnetic field intensity
 0  4 10 7 Wb Am

r 
Relative permeability
0
Ampère’s Law:
H

dl

i


Ampère’s Law
The line integral of the magnetic field
intensity around a closed path is equal to the
sum of the currents flowing through the
surface bounded by the path.
Magnetic Field Intensity and
Ampère’s Law
H  dl  Hdl cos( ) dot product
If the magnetic field H has a constant
magnitude and points in the same direction
as the incremental length dl Hl  i
Magnetic Circuits
In many engineering applications, we need to
compute the magnetic fields for structures that
lack sufficient symmetry for straight-forward
application of Ampère’s law. Then, we use an
approximate method known as magnetic-circuit
analysis.
magnetomotive force (mmf) of an N-turn
current-carrying coil
F NI
Analog: Voltage (emf)
reluctance of a path for magnetic flux

R 
A
Analog: Resistance
F  R
Analog: Ohm’s Law
Magnetic Circuit for Toroidal Coil
l  2R
A  r 2
1  l  1  2R  1  2 R 
R     2   2 
  A    r    r 
F  NI
F
 Nr I


R
2R
2
A Magnetic Circuit with Reluctances in
Series and Parallel
A Magnetic Circuit with Reluctances in
Series and Parallel
Rtotal  Rc 
1
1
1

Ra Rb
Ni
c 
Rtotal
a 
Rb
c
Ra  Rb
b 
Ra
c
Ra  Rb
Ba 
Ba 
a
Aa
a
Aa
(current divider)
Mutual Inductance
Self
inductance
for coil 1
Self
inductance
for coil 2
11
L1 
i1
11

i1
L2 

Mutual inductance between coils 1 and 2:
M
21
i1

21
i1

12
i2

12
i2
2 2
i2
22
i2
Mutual Inductance
Total fluxes linking the coils:
1  11  12
2  22  21
Mutual Inductance
Currents entering the dotted terminals
produce aiding fluxes
Circuit Equations for Mutual
Inductance
1  L1i1  Mi2
2   Mi1  L2i2
d1
di1
di2
e1 
 L1
M
dt
dt
dt
d2
di1
di2
e2 
 M
 L2
dt
dt
dt
Ideal Transformers
Ideal Transformers
N2
v 2 t  
v1 t 
N1
N1
i2 (t ) 
i1 (t )
N2
p2 t   p1 t 
Mechanical Analog
d1
d2
N2
v2 t  
v1 t 
N1
l2
d 2  d1
l1
N1
i2 (t ) 
i1 (t )
N2
l1
F2  F1
l2
Impedance Transformations
2
V1  N 1 
 Z L
Z L 
 
I1  N 2 
Semiconductor Diode
Shockley Equation
  vD
iD  I s exp 
  nVT
 
  1
 
 vD
iD  I s exp 
 nVT

 for v D  VT

kT
VT 
 26 mV
q
iD   I s for vD  -VT until reaching reverse breakdown
Load-Line Analysis of Diode Circuits
VSS  Ri D  v D
Assume VSS and R are known. Find iD and vD
Load-Line Analysis of Diode Circuits
VSS  Ri D  v D
VSS  Ri D  v D
Ideal Diode Model
The ideal diode acts as a short
circuit for forward currents
and as an open circuit with
reverse voltage applied.
iD > 0 vD < 0  diode is in the “on” state
vD < 0 ID = 0  diode is in the “off” state
Assumed States for Analysis of
Ideal-Diode Circuits
1. Assume a state for each diode, either on (i.e., a
short circuit) or off (i.e., an open circuit). For n
diodes there are 2n possible combinations of
diode states.
2. Analyze the circuit to determine the current
through the diodes assumed to be on and the
voltage across the diodes assumed to be off.
3. Check to see if the result is consistent with the
assumed state for each diode. Current must flow
in the forward direction for diodes assumed to
be on. Furthermore, the voltage across the
diodes assumed to be off must be positive at the
cathode (i.e., reverse bias).
4. If the results are consistent with the assumed
states, the analysis is finished. Otherwise, return
to step 1 and choose a different combination of
diode states.
Half-Wave Rectifier with Resistive Load
The diode is on during the positive half of the cycle. The
diode is off during the negative half of the cycle.
Half-Wave Rectifier with Smoothing
Capacitor
The charge removed from
the capacitor in one cycle:
Q  I LT  VC  Vr C
Full-Wave Rectifier
The capacitance required for a
full-wave rectifier is given by:
I LT
C
2Vr
Full-Wave Rectifier
Clipper Circuit
NPN Bipolar Junction Transistor
Bias Conditions for PN Junctions
The base emitter p-n
junction of an npn
transistor is normally
forward biased
The base collector p-n
junction of an npn
transistor is normally
reverse biased
Equations of Operation
  v BE
iE  I ES exp 
  VT
 
  1
 
From Kirchoff’s current law:
iE  iC  iB
Equations of Operation
Define  as the ratio of collector current to emitter
current:
iC

iE
Values for  range from 0.9 to 0.999 with 0.99 being
typical. Since:
iE  iC  iB  0.99iE  iB  iB  0.01iE
Most of the emitter current comes from the collector
and very little (1%) from the base.
Equations of Operation
Define  as the ratio of collector current to base
current:
iC

 
iB 1  
Values for  range from about 10 to 1,000 with a
common value being   100.
iC  iB
The collector current is an amplified version of the
base current.
The base region is very thin
Only a small fraction of the emitter current flows into the base
provided that the collector-base junction is reverse biased and the
base-emitter junction is forward biased.
Common-Emitter Characteristics
vBC
vCE
v BC  v BE  vCE
if v CE  v BE  v BC  0  reverse bias
Common-Emitter Input
Characteristics
  vBE
iB  (1   ) I ES exp 
  VT
 
  1
 
Common-Emitter Output
Characteristics
iC   iB for   100
Amplification by the BJT
A small change in vBE results in a large change in iB if the
base emitter is forward biased. Provided vCE is more than a
few tenth’s of a volt, this change in iB results in a larger
change in iC since iC=iB.
Common-Emitter Amplifier
Load-Line Analysis of a Common Emitter
Amplifier (Input Circuit)
VBB  vin t   RBiB t   vBE t 
Load-Line Analysis of a Common Emitter
Amplifier (Output Circuit)
VCC  RC iC  vCE
Inverting Amplifier
As vin(t) goes positive, the load line moves upward and to the
right, and the value of iB increases. This causes the operating
point on the output to move upwards, decreasing vCE  An
increase in vin(t) results in a much larger decrease in vCE so that
the common emitter amplifier is an inverting amplifier
PNP Bipolar Junction Transistor
Except for reversal of current directions and
voltage polarities, the pnp BJT is almost
identical to the npn BJT.
PNP Bipolar Junction Transistor
iC   i E
i B  (1   )i E
iC   i B
i E  iC  i B
NMOS Transistor
NMOS Transistor
Operation in the Cutoff Region
iD  0 for vGS  Vto
Operation Slightly Above Cut-Off
By applying a positive bias
between the Gate (G) and the body
(B), electrons are attracted to the
gate to form a conducting n-type
channel between the source and
drain. The positive charge on the
gate and the negative charge in the
channel form a capacitor where:
Operation Slightly Above Cut-Off
iDS 
W
Lt ox
(vGS  Vto )v DS
For small values of vDS, iD is proportional to vDS. The device
behaves as a resistance whose value depends on vGS.
Operation in the Triode Region

iD  C 2

2
vGS  vto vDS  vDS
 W  KP
C  
L 2

Load-Line Analysis of a Simple NMOS
Circuit
vGS (t )  vin (t )  4V
Load-Line Analysis of a Simple NMOS
Circuit
v DD  RDiD t   v DS t 
Load-Line Analysis of a Simple NMOS
Circuit
CMOS Inverter
Two-Input CMOS NAND Gate
Two-Input CMOS NOR Gate